Question

If $f\left(x\right)=\left|\begin{array}{ccc}{\left(1+x\right)}^a& {\left(1+2x\right)}^b& 1\\ 1& {\left(1+x\right)}^a& {\left(1+2x\right)}^b\\ {\left(1+2x\right)}^b& 1& {\left(1+x\right)}^a\end{array}\right|$ $a,b$ being positive integers, then

• A. Constant term or $f(x)$ is
• B. Coefficient of $x$ in $f(x)$ is 0
• C. Constant term in $f(x)$ is $a-b$
• D. Coefficient of $x$ in $f(x)$ is -1

Answer 1

Answer: (B)

let $\left|\begin{array}{ccc}{\left(1+x\right)}^a& {\left(1+2x\right)}^b& 1\\ 1& {\left(1+x\right)}^a& {\left(1+2x\right)}^b\\ {\left(1+2x\right)}^b& 1& {\left(1+x\right)}^a\end{array}\right|=a_0+a_{1}x+a_2{x}^2\dots$
Putting x = 0, we get
$a_0=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|=0$
Now differentiating both sides with respect to x and puttingx = 0, we get $a_1=\left|\begin{array}{ccc}a& 2b& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|+\left|\begin{array}{ccc}1& 1& 1\\ 0& a& 2b\\ 1& 1& 1\end{array}\right|+\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 2b& 0& a\end{array}\right|=0$
Hence, coefficient of $x$ is $0$